Sum of digits

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nikhtas30

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If \(\displaystyle k\in\mathbb{Z^+}\wedge 9k+1\), prove that

the reapiting sum digits of \(\displaystyle 9k+1\) equals at the end \(\displaystyle 1\)

For example if:
\(\displaystyle k=1\) then \(\displaystyle 9*1+1=10\rightarrow 1+0=1\)
\(\displaystyle k=2\) then \(\displaystyle 9*2+1=19\rightarrow 1+9=10\rightarrow 1+0=1\)
\(\displaystyle k=15\) then \(\displaystyle 9*15+1=136\rightarrow 1+3+6=10\rightarrow 1+0=1\)


Always it ends to 1. Why?
 
nikhtas30 said:
If \(\displaystyle k\in\mathbb{Z^+}\wedge 9k+1\), prove that

the reapiting sum digits of \(\displaystyle 9k+1\) equals at the end \(\displaystyle 1\)

For example if:
\(\displaystyle k=1\) then \(\displaystyle 9*1+1=10\rightarrow 1+0=1\)
\(\displaystyle k=2\) then \(\displaystyle 9*2+1=19\rightarrow 1+9=10\rightarrow 1+0=1\)
\(\displaystyle k=15\) then \(\displaystyle 9*15+1=136\rightarrow 1+3+6=10\rightarrow 1+0=1\)


Always it ends to 1. Why?
Click to expand...
Repeated sum of digits of 9*k = 9 → Repeated sum of digits of 9*k + 1= 10 → Repeated sum of digits of 9*k + 1= 1

(that is a method of calculating "divisibility" by 9)
 
Subhotosh Khan said:
Repeated sum of digits of 9*k = 9 → Repeated sum of digits of 9*k + 1= 10 → Repeated sum of digits of 9*k + 1= 1

(that is a method of calculating "divisibility" by 9)
Click to expand...

Thank you for your time.
I have another one question.
If we have for example:
\(\displaystyle k=11\) then \(\displaystyle 9*11+1=100\rightarrow 1+0+0=1\)
We didn't reach \(\displaystyle 10\) immediately. But we went to \(\displaystyle 1\).

So this thought: Repeated sum of digits of 9*k + 1= 10 Is it still true?
 
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nikhtas30 said:
If \(\displaystyle k\in\mathbb{Z^+}\wedge 9k+1\), prove that

the reapiting sum digits of \(\displaystyle 9k+1\) equals at the end \(\displaystyle 1\)

For example if:
\(\displaystyle k=1\) then \(\displaystyle 9*1+1=10\rightarrow 1+0=1\)
\(\displaystyle k=2\) then \(\displaystyle 9*2+1=19\rightarrow 1+9=10\rightarrow 1+0=1\)
\(\displaystyle k=15\) then \(\displaystyle 9*15+1=136\rightarrow 1+3+6=10\rightarrow 1+0=1\)


Always it ends to 1. Why?
Click to expand...
The basic idea is that when you add the digits of a number, the result differs from the number itself by a multiple of 9. This is because, for example, a 5 in the hundreds place of the number represents 5*100, while in the digit sum it is just 5; the difference between these is 5*(100-1) = 5*99 which is a multiple of 9. The same sort of thing happens with any digit in any place.

Therefore, each time you sum digits you are subtracting some multiple of 9 from it, and the remainder on division by 9 is unchanged.

When you repeat until you have a single digit, that digit is the remainder (unless it is 9 itself).

The number you are starting with has remainder 1; so the repeated digit sum must be 1.
 
I would prefer to see a strictly analytic algebric solution in order to understand it clearly.
 
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nikhtas30 said:
I would prefer to see a strictly analytic algebraic solution in order to understand it clearly.
Click to expand...
See if you can turn my argument into the kind you want! I find people grasp the idea more quickly from an example, so that's what I start with; symbolic explanations tend to get bogged down in details. But since you're happier with symbols, give it a try. That's a better way to learn than having it handed to you.

It's actually a lot easier if you can use number theory concepts and notations, such as modular arithmetic. Are you familiar with any of that?
 
I am not gonna lie, but I am not familiar with modular. I never used consciously modular and I never tried to solve exercises with modular before. But I can understand, what the symbols means. So if you prove it I will be happy.
 
nikhtas30 said:
I am not gonna lie, but I am not familiar with modular. I never used consciously modular and I never tried to solve exercises with modular before. But I can understand, what the symbols means. So if you prove it I will be happy.
Click to expand...
That doesn't surprise me, since you said "algebraic", not "number theory".

For an algebraic explanation, see https://en.wikipedia.org/wiki/Casting_out_nines

For number theoretic details, see the link there to digital root.

And if you are not familiar with casting out nines, a search for that term will bring you lots of information.
 
Dr.Peterson said:
That doesn't surprise me, since you said "algebraic", not "number theory".

For an algebraic explanation, see https://en.wikipedia.org/wiki/Casting_out_nines

For number theoretic details, see the link there to digital root.

And if you are not familiar with casting out nines, a search for that term will bring you lots of information.
Click to expand...
Thank you, Thank you, Thank you, Thank you... Very much for your time and help Thank you!
 
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